I know that being connected is a topological property. But, my question is what are examples of two spaces where this isn't helpful?

Is this simply two spaces that are disconnected or having some other topological property? I know that that is also a topological property, but I'm asking as if connected is the only property known at this point - for the purposes of this question.

Moreover, why are so many different topological properties needed / used. Is this simply due to the wide array of topologies there can be?

  • $\begingroup$ Any property will not help with distinguishing two things if both of them have it or both of them don't have it. This is true for anything, not just topological spaces. Are you asking about examples of pairs of spaces which are connected, but not homeomorphic? $\endgroup$
    – tomasz
    Dec 13, 2017 at 2:45
  • $\begingroup$ @tomasz Yes, sorry could have been clearer about this. $\endgroup$ Dec 13, 2017 at 2:45
  • $\begingroup$ A topology is indeed a very general concept. The spaces $A=(0,1)$ and $B= [0,1)$ are connected and not compact and share many topological properties, but they are not homeomorphic, because if $p$ is any member of $A$ then the subspace $A$ \ $\{p\}$ is not a connected subspace of $A$, while $B$ \ $\{0\}$ is a connected subspace of $B$. $\endgroup$ Dec 13, 2017 at 3:07

3 Answers 3


Here is just a bag-grab:

homology: Consider $\mathbb R^2$ and $\mathbb R^3$. These are not homeomorphic, but they are both connected, and neither have cut points.

Most of the tools for showing two objects are not homeomorphic comes from showing that they are not even homotopy equivalent. For this property, we can assign algebraic invariants, in this case homology does the trick.

cut points: Another example is usually $S^1$ and $[0,1]$. These are connected, but if there were a homeomorphism $\phi:S^1 \to [0,1]$, then by surjectivity, there is some point $a$ that maps to $.5 \in [0,1]$, so the homeomorphism should restrict to $S^1\setminus\{a\} \to [0,1]\setminus\{.5\}$, but the latter is disconnected. This is a usual argument by cut points.

compactness: $(0,1)$ is not homeomorphic to $[0,1]$, since one is compact and one is not.

Cardinality: $\mathbb Z$ is not homeomorphic to $(n,n+1)$ with $n \in \mathbb Z$ under any topologies since they have different cardinalities (but are both really disconnected.)

There are other variations on a theme: there are linking numbers for knots, all kinds of local connectedness properties etc.

  • $\begingroup$ Hmmm. I might have misunderstood your answer or badly worded my question. More specifically, "what are two spaces which can be shown to not be homeomorphic by using connectedness as a topological property." Not sure if your answer still applies with this rewording. $\endgroup$ Dec 13, 2017 at 2:48
  • $\begingroup$ In that case, just take two spaces with a different number of components. $\endgroup$
    – anomaly
    Dec 13, 2017 at 2:49
  • $\begingroup$ @Overflow2341313 you answered the question in OP by saying "I'm asking for spaces that are both connected but not homeomorphic." The final line of your question asks "why are so many different topological properties needed / used" $\endgroup$ Dec 13, 2017 at 2:58
  • $\begingroup$ What do you want to know? Why connectedness is useful? What other things there are out there? $\endgroup$ Dec 13, 2017 at 2:59

Similar to Andres' answer, you can show that $\mathbb R$ and $\mathbb R^2$ are not homeomorphic by the preservation of connected components. Indeed, if there was a homeomorphism $f:\mathbb R \to \mathbb R^2$, then the restriction $f:\mathbb R\setminus\left\{0\right\} \to \mathbb R^2\setminus\left\{f(0)\right\}$ should also be a homeomorphism. But $\mathbb R\setminus \left\{0\right\}$ is not connected, while $\mathbb R^2$ is (path) connected.

  • $\begingroup$ Sorry, the question was changed while I was writing. I am not sure this is what you are looking for. $\endgroup$
    – Ben Tighe
    Dec 13, 2017 at 2:59

In order to prove that two spaces $X$ and $Y$ are homeomorphic you have to present a bijective map $f:\>X\to Y$ which is continuous in both directions. Showing that certain topological properties or invariants coincide is not enough, unless both spaces are members of a well understood family, e.g., "closed surfaces" or "open regions in ${\mathbb R}^2$".

In order to prove that two spaces $X$ and $Y$, e.g., two knots in ${\mathbb R}^3$, are not homeomorphic you have to invoke properties or invariants that are different for the two spaces. There might be googols of bijective maps between $X$ and $Y$, and it would take an eternity to prove that each of these has a discontinuity somewhere.


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